What are the solutions to the equation -x^2-0x+16=0 ?

The solutions to the equation -x^2-0x+16=0 are x=-4,x=4

Find the zeros of -x^2-0x+16=0

The zeros of -x^2-0x+16=0 are x=-4,x=4

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- x^{2} - 0 x + 16 = 0

x = - 4, x = 4

Solution steps

- x^{2} - 0 \cdot x + 16 = 0

Refine

- x^{2} + 16 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 ( - 1 ) \cdot 16}{2 ( - 1 )}

0^{2} - 4 (- 1) \cdot 16 = 8

x_{1, 2} = \frac{- 0 \pm 8}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 0 + 8}{2 ( - 1 )}, x_{2} = \frac{- 0 - 8}{2 ( - 1 )}

x = \frac{- 0 + 8}{2 ( - 1 )} : - 4

x = \frac{- 0 - 8}{2 ( - 1 )} : 4

The solutions to the quadratic equation are:

x = - 4, x = 4

2 p^{2} + 2 p - 84 = 0

y, x^{2} y^{2} + x y = 12

3 y^{2} + 7 y - 10 = 0

30 + 13 x - 3 x^{2} = 0

4 (45 + x^{2}) = 196

What are the solutions to the equation -x^2-0x+16=0 ?
The solutions to the equation -x^2-0x+16=0 are x=-4,x=4
Find the zeros of -x^2-0x+16=0
The zeros of -x^2-0x+16=0 are x=-4,x=4